python - 遍历字符串并打印字符之间的距离

Question

source = 'abc'

def editDistance(source, target):
    items1=[]
    for c in range(0,len(source)):
        for k in range(1,len(source)):
            if (k < len(source)):
                test = ord(source[k]) - ord(source[c])
                items1.append(test)    
    return items1

我试图遍历字符串并找到字母表中每个字符之间的距离。因此，a 和 b 之间的距离是 1，b 和 c 之间的距离是> 是 1。我想打印出 [1, 1] 数组，但是，我认为我弄乱了 for 循环及其打印输出:[1, 2, 0, 1 , -1, 0].

Answer 1

这不就是这么简单吗:

alphabet = 'abcdefghijklmnopqrstuvwxyz'
abs(alphabet.index('z') - alphabet.index('a'))

这是一个概念证明:

Python 3.7.4 (default, Aug 12 2019, 14:45:07) 
[GCC 9.1.1 20190605 (Red Hat 9.1.1-2)] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> alphabet = 'abcdefghijklpmnopqrstuvwxyz'
>>> abs(alphabet.index('z') - alphabet.index('a'))
26
>>> abs(alphabet.index('a') - alphabet.index('c'))
2
>>> abs(alphabet.index('c') - alphabet.index('a'))
2
>>> 

它实际上适用于任何字符集，无论什么情况或类别:

Python 3.7.4 (default, Aug 12 2019, 14:45:07) 
[GCC 9.1.1 20190605 (Red Hat 9.1.1-2)] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> s='ABCdefgh!çõ'
>>> abs(s.index('ç') - s.index('A'))
9
>>> abs(s.index('B') - s.index('A'))
1
>>> abs(s.index('B') - s.index('!'))
7
>>> abs(s.index('!') - s.index('B'))
7
>>>